Post on 19-Oct-2015
Ken Youssefi Mechanical Engineering Dept., SJSU 1
Failure Theories Static Loads
Static load a stationary load that is gradually applied having an unchanging magnitude and direction
Failure A part is permanently distorted and will not function properly.
A part has been separated into two or more pieces (fracture)
Material Strength
Sy = Yield strength in tension, Syt = Syc
Sys = Yield strength in shear
Su = Ultimate strength in tension, Sut
Suc = Ultimate strength in compression
Sus = Ultimate strength in shear = .67 Su
Ken Youssefi Mechanical Engineering Dept., SJSU 2
Ductile and Brittle Materials
A ductile material deforms significantly before fracturing. Ductility is measured
by % elongation at the fracture point. Materials with 5% or more elongation are
considered ductile.
Brittle material yields very little before fracturing, the
yield strength is approximately the same as the
ultimate strength in tension. The ultimate strength in
compression is much larger than the ultimate
strength in tension.
Ken Youssefi Mechanical Engineering Dept., SJSU 3
Failure Theories Ductile Materials
Maximum shear stress theory (Tresca 1886)
Yield strength of a material is used to design components made of
ductile material
= Sy
Sy
2 =
= Sy
=Sy
(max )component > ( )obtained from a tension test at the yield point Failure
(max )component < Sy
2
To avoid failure
max = Sy
2 n n = Safety factor
Design equation
Ken Youssefi Mechanical Engineering Dept., SJSU 4
Failure Theories Ductile Materials
Distortion energy theory (von Mises-Hencky)
t
t
Simple tension test (Sy)t
(Sy)t (Sy)h >>
Distortion contributes to
failure much more than
change in volume.
(total strain energy) (strain energy due to hydrostatic stress) = strain energy
due to angular distortion > strain energy obtained from a tension test at the yield point failure
Hydrostatic state of stress (Sy)h
h
h
h
Ken Youssefi Mechanical Engineering Dept., SJSU 5
Failure Theories Ductile Materials
The area under the curve in the elastic region is called the Elastic Strain Energy.
Strain
energy
U =
3D case
UT = 11 + 22 + 33
1 = 1 E
2 E
3 E
v v
2 = 2 E
1 E
3 E
v v
3 = 3 1
E
2 E
v v
Stress-strain relationship
E
UT = (12 + 2
2 + 32) - 2v (12 + 13 + 23)
2E
1
Ken Youssefi Mechanical Engineering Dept., SJSU 6
Failure Theories Ductile Materials
Ud = UT Uh
Distortion strain energy = total strain energy hydrostatic strain energy
Substitute 1 = 2 = 3 = h
Uh = (h2 + h
2 + h2) - 2v (hh + hh+ hh)
2E
1
Simplify and substitute 1 + 2 + 3 = 3h into the above equation
Uh = (1 2v) = 2E
3h2
6E
(1 2v) (1 + 2 + 3)2
UT = (12 + 2
2 + 32) - 2v (12 + 13 + 23)
2E
1 (1)
Ud = UT Uh = 6E
1 + v (1 2)
2 + (1 3)2 + (2 3)
2
Subtract the hydrostatic strain energy from the total energy to
obtain the distortion energy
(2)
Ken Youssefi Mechanical Engineering Dept., SJSU 7
Failure Theories Ductile Materials
Strain energy from a tension test at the yield point
1= Sy and 2 = 3 = 0 Substitute in equation (2)
3E
1 + v (Sy)
2 Utest =
To avoid failure, Ud < Utest
(1 2)2 + (1 3)
2 + (2 3)2
2
< Sy
Ud = UT Uh = 6E
1 + v (1 2)
2 + (1 3)2 + (2 3)
2 (2)
Ken Youssefi Mechanical Engineering Dept., SJSU 8
Failure Theories Ductile Materials
2D case, 3 = 0
= (12 12 + 2
2) < Sy Where is von Mises stress
= Sy
n Design equation
(1 2)2 + (1 3)
2 + (2 3)2
2
< Sy
Ken Youssefi Mechanical Engineering Dept., SJSU 9
Pure torsion, = 1 = 2
(12 2 1 + 2
2) = Sy2
Failure Theories Ductile Materials
32 = Sy
2 Sys = Sy / 3 Sys = .577 Sy
Relationship between yield strength in
tension and shear
(x)2 + 3(xy)
2 =
Sy
n
1/2
If y = 0, then 1, 2 = x/2 [(x)/2]2 + (xy)
2
the design equation can be written in terms of the dominant
component stresses (due to bending and torsion)
Ken Youssefi Mechanical Engineering Dept., SJSU 10
Design Process
= Sy
n max =
Sy 2n
Maximum shear stress theory Distortion energy theory
consider strength, environment, cost, weight,
availability, codes and standards, manufacturing process Sy , Su
Choose a safety factor
The selection of an appropriate safety factor should be based on the following:
Degree of uncertainty about loading (type, magnitude and direction)
Degree of uncertainty about material strength
Type of manufacturing process
Uncertainties related to stress analysis
Consequence of failure; human safety and economics
Codes and standards
n Cost Weight Size
Select material:
Ken Youssefi Mechanical Engineering Dept., SJSU 11
Design Process
Use n = 1.2 to 1.5 for reliable materials subjected to
loads that can be determined with certainty.
Use n = 1.5 to 2.5 for average materials subjected to
loads that can be determined. Also, human safety and
economics are not an issue.
Use n = 3.0 to 4.0 for well known materials subjected to
uncertain loads.
Ken Youssefi Mechanical Engineering Dept., SJSU 12
Design Process - Static load & Ductile material
Formulate the maximum stresses in the component in terms of size: x, xy
Optimize for weight, size, or cost.
Select material Sy , Su
Choose a safety factor, n
Determine the principal stresses, maximum shear stress and von
Mises stress in terms of the size; 1, 2 , max and Use appropriate failure theory to calculate the size.
= Sy
n max =
Sy 2n
Distortion energy theory Maximum shear stress theory
Choose a cross section: round, rectangular, hollow, I-beam, C channel,
Ken Youssefi Mechanical Engineering Dept., SJSU 13
Failure Theories Brittle Materials
Characteristics of brittle materials;
Perform two tests, one in compression and one in tension, draw the
Mohrs circles for both tests.
Compression test
Suc
Failure envelope
The component is safe if the state of stress falls inside the failure envelope.
1 > 2 and 3 = 0
Tension test
Sut 2 1 Stress state
2. Suc >> Sut 1. Sut Syt 3. elongation < 5%
Ken Youssefi Mechanical Engineering Dept., SJSU 14
Failure Theories Brittle Materials
1 Sut
Suc
Sut
Suc
Safe
Safe
Safe Safe
-Sut
Cast iron data
Modified Coulomb-Mohr theory
1
2 or 3
Sut
Sut
Suc
-Sut
I
II
III
Three design zones
2 or 3
Ken Youssefi Mechanical Engineering Dept., SJSU 15
Failure Theories Brittle Materials
1
2
Sut
Sut
Suc
-Sut
I
II
III
Zone I
1 > 0 , 2 > 0 and 1 > 2
Zone II
1 > 0 , 2 < 0 and 2 < Sut
Zone III
1 > 0 , 2 < 0 and 2 > Sut
1 ( 1
Sut
1
Suc )
2 Suc
= 1
n
Design equation
1 = Sut
n Design equation
1 = Sut
n Design equation